# A simple demonstration that the electrostatic potential has no extrema in free space

The electrostatic potential $V$ cannot attain a local maximum or minimum in free space (a region of space where there are no charges).

The claim makes sense intuitively: if $V$ had a local maximum at some point $P$, then since field lines move from higher to lower potential, the field lines near $P$ would all be moving away from $P$, i.e the divergence of the field would be positive at $P$ or said otherwise $P$ would be acting like a source of the field. But the only sources of the electric field (in the electrostatic approximation) are charges, so there would have to be a charge at $P$.

A mathematically rigorous way: the electrostatic potential verifies Poisson's equation $\nabla^2 V=-\rho/\epsilon_0$, but in free space $\rho=0$, so $V$ satisfies Laplace's equation: $\nabla^2 V=0$ (i.e $V$ is harmonic). But harmonic functions satisfy have no local max or min.

But the above is rather mathematical advanced. Is there some way to prove the claim without relying on high powered tools such as the maximum principle for harmonic functions.

• This is Earnshaw's theorem, you could look up that proof. It's essentially just making your intuitive argument rigorous using Gauss's law in integral form. – QtizedQ Jul 11 '17 at 23:22
• Is it just that, to satisfy $\nabla^2V = 0$, one of the second derivatives must have the opposite sign of the others, while, at a max/min, they must have the same sign? – Alfred Centauri Jul 11 '17 at 23:41
• You can exploit the fact that in free space, the average potential over the surface of some (imaginary) sphere equals the potential at the sphere's center. If you were to center the sphere at a local min/max and then change its radius, the surface-averaged potential will depend on chosen radius ergo a local min/max cannot exist. I believe this is called the "Mean Value Theorem" in some cases. – forky40 Jul 12 '17 at 0:15

The maximum principle for harmonic functions is actually based on a really simple and intuitive idea: for a harmonic function, the value of $V$ at a point $V$ is really just the average value that $V$ takes on neighboring points. This can be seen by looking at the problem on a lattice of points (let's stay in $2D$ for now). If we label our points as $x_{ij}$, then the first "derivative" looks like

$$\boldsymbol{\nabla}V=\frac{1}{a}\left(V(x_{i+1/2,j})-V(x_{i-1/2,j})\right)\hat{\textbf{x}}+\frac{1}{a}\left(V(x_{i,j+1/2})-V(x_{i,j-1/2}))\right)\hat{\textbf{y}}$$

(note that since this uses indices like $i\pm 1/2$, the gradient actually isn't well defined -- however, the Laplacian will be.) The "gradient" of this gives you the lattice Laplacian:

$$\nabla^2V=\frac{1}{a^2}\left(V(x_{i+1,j})+V(x_{i-1,j})+V(x_{i,j+1})+V(x_{i,j-1})-4V(x_{ij})\right).$$

Setting this equal to zero tells you that

$$V(x_{ij})=\frac{1}{4}\left(V(x_{i+1,j})+V(x_{i-1,j})+V(x_{i,j+1})+V(x_{i,j-1})\right),$$

so that $V(x)$ is really just an average of $V$ at neighboring points! This continues to hold in the $a\to 0$ limit, but let's stick with the discretized version to see what we can get.

Firstly, this argument shows very easily that a function which is harmonic on some domain $\Sigma$ must take its maximum on the boundary $\partial\Sigma$. If this weren't true, then it would have to be true that a point $x$ where $V(x)$ is maximal would have to be the average of points which are not maximal. That was easy! Applying the argument to $-V$ shows the same thing for the minima.

Secondly, it becomes very easy to show that solutions are unique. First, note that if a function is harmonic on $\Sigma$ and satisfies $V=0$ on $\partial\Sigma$, then $V=0$ everywhere on $\Sigma$. If not, then it would have its maximum on the interior of $\Sigma$, contradicting the above statement. Now, consider two harmonic functions $V_1$ and $V_2$ on $\Sigma$ which take the same values on the boundary $\partial\Sigma$. Then their difference $V_3=V_2-V_1$ is also harmonic and takes the value $V_3=0$ on the boundary. Thus, $V_3=0$ on $\Sigma$, so that $V_1=V_2$, showing that solutions are unique once boundary conditions are fixed. Thus, taking the continuum limit $a\to 0$, you are given your conclusions!

Another way to see this is to consider a point $x_0$ and then consider the average value of $V$ on a sphere of radius $R$ with $x_0$ at its center (let's take $x_0$ to be the origin for simplicity). This is given by

$$\frac{1}{4\pi R^2}\oint_{S_R}V(\textbf{x})\,\mathrm{d}S=\frac{1}{4\pi R^3}\oint_{S_R}(V(\textbf{x})\,\textbf{x})\cdot\mathrm{d}\textbf{S}=\frac{1}{4\pi}\oint_{S_R}V(\textbf{x})\frac{\textbf{x}}{|\textbf{x}|^3}\cdot\mathrm{d}\textbf{S}\\ =-\frac{1}{4\pi}\oint_{S_R}\boldsymbol{\nabla}\left(V(\textbf{x})\frac{1}{|\textbf{x}|}\right)\cdot\mathrm{d}\textbf{S}+\frac{1}{4\pi}\oint_{S_R}\frac{1}{|\textbf{x}|}\boldsymbol{\nabla}V(\textbf{x})\cdot\mathrm{d}\textbf{S}\\ =-\frac{1}{4\pi}\int_{B_R}\nabla^2\left(\frac{V(\textbf{x})}{|\textbf{x}|}\right)\mathrm{d}V,$$

where the second term in the second line vanishes by the divergence theorem and the fact that $\nabla^2V=0$. Now, noting that

$$\nabla^2\left(\frac{1}{|\textbf{x}|}\right)=-4\pi\delta(\textbf{x}),$$

and using traditional product rules for the Laplacian, we can obtain

$$\frac{1}{4\pi R^2}\oint_{S_R}V(\textbf{x})\,\mathrm{d}S=V(0),$$

which is a more rigorous derivation of the maximum principle.

I hope this helps!

You don't really need to appeal to the theory of harmonic functions. If $V$ has an extremum at some point, then its gradient is zero and its Hessian is positive/negative definite, i.e., its eigenvalues are either all positive or all negative. But the Laplacian is the trace of the Hessian matrix, i.e., the sum of the eigenvalues, and in free space this is zero. Therefore the Hessian can't be positive or negative definite.

• This is wrong, $V(\mathbf{x})=|\mathbf{x}|^4$ has a (global) minium in zero, where the Hessian vanishes. – fqq May 4 at 13:08